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I've been trying to understand how to analyse circuits with transistors, but I'm having a hard time finding out where to start when confronted with the following two types of circuits. (Gain = infinity)

In the first circuit I've tried applying kirchhoff's laws to find out some currents and voltages but I got stuck finding the current flowing through the diode and the voltage Vce.

First Circuit

In the second circuit I simply can't start doing anything, I don't know what to do with the current source and how to start.

Second Circuit

I don't want the solution, but some guidance(as where to start, how to approach these situations) would be really much appreciated. I'm really sorry if this post asks some stupid questions.

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    \$\begingroup\$ Gain = infinity Neither of these circuits have an infinite gain. Perhaps you mean that the \$\beta\$ of the NPN transistors can be considered infinite. You should have had some education on how to analyze circuits like this. Without that it is simply impossible to analyze them. Start by assuming that the \$V_{BE}\$ of the NPNs is 0.7 V, also the diode has a forward voltage of 0.7 V, where does that get you? \$\endgroup\$ – Bimpelrekkie Jan 23 '20 at 13:54
  • \$\begingroup\$ Sorry for the "gain" but I didn't know how to name it as english isn't my first language and after looking at some posts about circuit analysis with transistors and seeing it called "Gain" I assumed that that's what they talked about. I already assumed that Vdiode is 0.6 and Vbe is 0.6 but after applying 2nd Kirchoff law on almost every loop I can't find a connection. In the first circuit I'm more interested in what impact would make the current source I on the circuit. There (Vb node) I tried to apply 1st kirchoff law but I can't seem to make a connection with these. \$\endgroup\$ – unhappy_b Jan 23 '20 at 14:10
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    \$\begingroup\$ These are instructive circuits. In ordinary - voltage supply only - bias circuits one can usually manage to walk through the circuit solving at most one equation at the time. With these current source 'tainted' circuits you will end up having to solve at least two equations at once. After assuming Vbe and Vd to be 0.6V, add all voltage drops on resistors that are known due to the fact you know the current in them. The second circuit seems even easier to solve. (Also, remember that the current sources will provide whatever voltage is required to supply their given current). \$\endgroup\$ – Sredni Vashtar Jan 23 '20 at 15:21
  • \$\begingroup\$ @unhappy_b Are you allowed to use fixed voltage drop assumptions for Vbe and Vd in the first circuit? (Because that's not how they work.) \$\endgroup\$ – jonk Jan 23 '20 at 15:23
  • \$\begingroup\$ Using fixed Vbe and Vd, and also neglecting Ib will give round numbers :-) The result can be used as a starting point for a refinement. \$\endgroup\$ – Sredni Vashtar Jan 23 '20 at 15:29
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First Circuit

Let's look at the base voltage of the first circuit using what I believe to be your stated facts:

  • \$I_\text{B}=0\:\text{A}\$ (\$\beta\to \infty\$)
  • \$V_\text{BE}=600\:\text{mV}\$
  • \$V_\text{D}=600\:\text{mV}\$
  • \$V_\text{CC}=5.6\:\text{V}\$

Swapping the diode near the base with its series resistor, without damage to the analysis, we can now write using KCL:

$$\begin{align*}I+\frac{V_\text{B}}{R_{\text{B}_1}}+\frac{V_\text{B}}{R_{\text{B}_2}}+I_\text{B}&=\frac{V_\text{D}}{R_{\text{B}_1}}+\frac{V_\text{CC}}{R_{\text{B}_2}}\\\\1\:\text{mA}+\frac{V_\text{B}}{1\:\text{k}\Omega}+\frac{V_\text{B}}{1\:\text{k}\Omega}+0\:\text{A}&=\frac{5.6\:\text{V}}{1\:\text{k}\Omega}+\frac{600\:\text{mV}}{1\:\text{k}\Omega}\\\\&\therefore\\\\V_\text{B}&=2.6\:\text{V}\\\\V_\text{E}&=2.0\:\text{V}=V_\text{B}-V_\text{BE}\\\\V_\text{C}&=4.6\:\text{V}=V_\text{CC}-I_2\cdot R_\text{C}\\\\V_\text{CE}&=2.6\:\text{V}=V_\text{C}-V_\text{E}\\\\I_\text{D}&=2\:\text{mA}=\frac{V_\text{B}-V_\text{D}}{R_{\text{B}_1}}\end{align*}$$

Second Circuit

Since you only want guidance, for now, I'd suggest the following steps in the indicated order:

  1. Label the top node as the unknown value \$V_\text{CC}\$.
  2. Write down the nodal equation for \$V_\text{CC}\$.
  3. Write down the nodal equation for \$V_\text{B}\$.
  4. Write down the nodal equation for \$V_\text{E}\$.
  5. Write down the equation relating \$V_\text{B}\$ to \$V_\text{E}\$.
  6. Write down the equation relating \$V_\text{CC}\$ to \$V_\text{C}\$.
  7. Solve the above 5 equations simultaneously for these five unknowns: \$V_\text{CC}\$, \$V_\text{B}\$, \$V_\text{E}\$, \$V_\text{C}\$, and \$I_\text{D}\$.

(Keep in mind that you know the value for \$V_\text{D}\$, \$V_\text{BE}\$, and \$I_\text{CC}\$.)

As a hint you may use to verify things, the resulting values are as easy to write down as the answers found for the first circuit.

;)

Since someone decided to mark this question down (I don't mind, because it gives me fun excuses), I think I'll just write out the complete solution for you. You got lucky!

Using sympy:

var('vcc vc ve vb rb1 rb2 rc re id icc vbe vd')
eq1=Eq(vb/rb1+vb/rb2,vcc/rb2)
eq2=Eq(vcc/rb2+vcc/rc+id,icc+vc/rc+vb/rb2)
eq3=Eq(ve/re,(vcc-vc)/rc+id)
eq4=Eq(vb,ve+vbe)
eq5=Eq(vc,vcc-vd)
ans=solve([eq1,eq2,eq3,eq4,eq5],[vcc,vb,ve,vc,id])
for x in ans:x,ans[x].subs({vd:.6,vbe:.6,rb1:800,rb2:2000,re:1000,rc:1200,icc:3e-3})
(vb, 1.60000000000000)
(id, 0.000500000000000000)
(vc, 5.00000000000000)
(vcc, 5.60000000000000)
(ve, 1.00000000000000)

It solves out.

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  • \$\begingroup\$ Thank you for the detailed answer and the solution. I'm marking this as solved! But there's one more thing, in my solution you said that I'm ignoring the base current; shouldn't I ignore it? As long as it's 0? Thank you again for the help! \$\endgroup\$ – unhappy_b Jan 25 '20 at 13:03
  • \$\begingroup\$ @unhappy_b ignoring base current is an approximation that lead to much much easier calculations (and in fact it also leads to round numbers). If you simulate the circuit (and I encourage you to download the now free MicroCap 12 simulator) you will see that when you account for base current (by using a beta=200 transistor) the values found differ by about 5% with respect to the case of infinite beta. With base current you need to solve multiple equations simultaneously, with zero Ib, you have only one instance where you need to consider two equations at once, and can solve it right away. \$\endgroup\$ – Sredni Vashtar Jan 25 '20 at 13:36
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

  • $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$
  • Transistor gain \$\beta\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$

Now, using KCL, we can write:

$$ \begin{cases} \text{I}_\text{x}=\text{I}_3+\text{I}_4\\ \\ \text{I}_3=\text{I}_1+\text{I}_\text{t}+\text{I}_\text{d}\\ \\ \text{I}_2=\text{I}_\text{x}+\text{I}_5\\ \\ 0=\text{I}_\text{d}+\text{I}_1+\text{I}_5\\ \\ \text{I}_2=\text{I}_\text{t}+\text{I}_4\\ \\ \beta=\frac{\text{I}_4}{\text{I}_\text{t}} \end{cases}\tag3 $$

Using KVL, we get:

$$ \begin{cases} \text{I}_3=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{x}-\text{V}_5}{\text{R}_4}\\ \\ \text{I}_2=\frac{\text{V}_3-\text{V}_4}{\text{R}_2}\\ \\ \text{I}_\text{d}=\frac{\text{V}_2}{\text{R}_1}\\ \\ \text{V}_1-\text{V}_3=\alpha\\ \\ \text{V}_1-\text{V}_2=\eta \end{cases}\tag4 $$

Solving the circuit using your given values, and using \$\beta=10^{20}\$ and \$\alpha=\eta=\frac{7}{10}\$ leads to:

FullSimplify[
 Solve[{Ix == I3 + I4, 
   I3 == (1*10^(-3)) + It + Id, (1*10^(-3)) == Ix + I5, 
   0 == Id + (1*10^(-3)) + I5, (1*10^(-3)) == It + I4, 10^(20) == I4/It, 
   I3 == ((56/10) - V1)/1000, 
   I4 == ((56/10) - V5)/1000, (1*10^(-3)) == (V3 - V4)/1000, 
   Id == V2/1000, V1 - V3 == 7/10, V1 - V2 == 7/10}, {Id, It, Ix, I3, 
   I4, I5, V1, V2, V3, V4, V5}]]

So:

{{Id -> 3900000000000000000029/2000000000000000000020000, 
  It -> 1/100000000000000000001000, 
  Ix -> 7900000000000000000069/2000000000000000000020000, 
  I3 -> 5900000000000000000069/2000000000000000000020000, 
  I4 -> 100000000000000000/100000000000000000001, 
  I5 -> -(5900000000000000000049/2000000000000000000020000), 
  V1 -> 5300000000000000000043/2000000000000000000020, 
  V2 -> 3900000000000000000029/2000000000000000000020, 
  V3 -> 3900000000000000000029/2000000000000000000020, 
  V4 -> 1900000000000000000009/2000000000000000000020, 
  V5 -> 2300000000000000000028/500000000000000000005}}

And:

{{Id -> 0.00195, It -> 1.*10^-23, Ix -> 0.00395, I3 -> 0.00295, 
  I4 -> 0.001, I5 -> -0.00295, V1 -> 2.65, V2 -> 1.95, V3 -> 1.95, 
  V4 -> 0.95, V5 -> 4.6}}

EDIT 1

For \$\beta\to\infty\$:

{{Id -> 39/20000, It -> 0, Ix -> 79/20000, I3 -> 59/20000, 
  I4 -> 1/1000, I5 -> -(59/20000), V1 -> 53/20, V2 -> 39/20, 
  V3 -> 39/20, V4 -> 19/20, V5 -> 23/5}}

Wich is:

{{Id -> 0.00195, It -> 0., Ix -> 0.00395, I3 -> 0.00295, I4 -> 0.001, 
  I5 -> -0.00295, V1 -> 2.65, V2 -> 1.95, V3 -> 1.95, V4 -> 0.95, 
  V5 -> 4.6}}

EDIT 2

FullSimplify[
 Solve[{Ix == I3 + I4, I3 == I1 + It + Id, I2 == Ix + I5, 
   0 == Id + I1 + I5, I2 == It + I4, β == I4/It, 
   I3 == (Vx - V1)/R3, I4 == (Vx - V5)/R4, I2 == (V3 - V4)/R2, 
   Id == V2/R1, V1 - V3 == α, V1 - V2 == η}, {Id, It, Ix, 
   I3, I4, I5, V1, V2, V3, V4, V5}]]

Which gives:

{{Id -> -((I1 R3 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3)), 
  It -> I2/(1 + β), 
  Ix -> I2 - (I2 R3)/((R1 + R3) (1 + β)) + (
    I1 R1 + Vx - η)/(R1 + R3), 
  I3 -> (I2 R1 + 
    I1 R1 (1 + β) + (1 + β) (Vx - η))/((R1 + 
      R3) (1 + β)), I4 -> (I2 β)/(1 + β), 
  I5 -> (-I1 R1 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3), 
  V1 -> (-I2 R1 R3 - 
    I1 R1 R3 (1 + β) + (1 + β) (R1 Vx + R3 η))/((R1 +
       R3) (1 + β)), 
  V2 -> (R1 (-I1 R3 + Vx - (I2 R3)/(1 + β) - η))/(R1 + R3),
   V3 -> -α - (I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V4 -> -I2 R2 - α - (
    I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V5 -> Vx - (I2 R4 β)/(1 + β)}}
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  • \$\begingroup\$ beta = 10^20 and beta=infinity are nearly the same thing. How about you redo the first calculation with beta = 200? Or, better yet, why don't you make it parametric and show how Ic varies with beta? \$\endgroup\$ – Sredni Vashtar Jan 25 '20 at 13:43
  • \$\begingroup\$ @SredniVashtar Well, the OP can do that using my code ;) \$\endgroup\$ – Jan Jan 25 '20 at 13:53
  • \$\begingroup\$ Maybe he does not have MMA, or a Raspi with a free -slooooow- copy of it. :-) Seriously, it would be instructive to see how the approximation gets better for higher values of beta, or the other way around: when the approximation ceases to be a good approximation. \$\endgroup\$ – Sredni Vashtar Jan 25 '20 at 13:59
  • \$\begingroup\$ @SredniVashtar Done. \$\endgroup\$ – Jan Jan 25 '20 at 14:05
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    \$\begingroup\$ +1 for the effort (even if I do not favor this way of solving the circuit by writing all equations at once - I prefer a visual step-by-step walkthrough). So now we can see - not surpisingly - that Ic = I2 beta / (beta +1) and the percent error with respect to the ideal case of infinite beta (that is Ic = I2) can be visualized as a function of beta. \$\endgroup\$ – Sredni Vashtar Jan 25 '20 at 16:41
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For the first circuit(with VBE=VD=0.6 and β->∞) I used KCL and KVL to find the currents IRB1 and IRB2 as follows:

Equation

Next I used KVL to find VB,VE,VC, after that I found VCE:

enter image description here

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  • \$\begingroup\$ Thanks. So you really are ignoring the base current. Can you provide some computed values here, as well? \$\endgroup\$ – jonk Jan 24 '20 at 18:53

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